Problem Statement
Let $A\subseteq \mathbb{N}$ be an infinite arithmetic progression and $f:A\to \{-1,1\}$ be a non-constant function. Must there exist a finite non-empty $S\subset A$ such that\[\sum_{n\in S}\frac{f(n)}{n}=0?\]What about if $A$ is an arbitrary set of positive density? What if $A$ is the set of squares excluding $1$?
Categories:
Number Theory Unit Fractions
Progress
Erdős and Straus [ErSt75] proved this when $A=\mathbb{N}$. Sattler [Sa75] proved this when $A$ is the set of odd numbers. For the squares $1$ must be excluded or the result is trivially false, since\[\sum_{k\geq 2}\frac{1}{k^2}<1.\]This is false for some sets $A$ of positive density - indeed, it fails for any set $A$ containing exactly one even number. (Sattler [Sa82] credits this observation to Erdős, who presumably found this after [ErGr80].)Sattler [Sa82b] proved the answer to the original question is yes, in that any arithmetic progression has this property.
The final question of the set of squares excluding $1$ appears to be open - Sattler announced a proof in [Sa82] and [Sa82b], but this never appeared.
Source: erdosproblems.com/318 | Last verified: January 14, 2026